Should I use some laws of large numbers or should I refer to other technics? In a lecture (probability theory) note, there is a starred exercise in the chapter of convergence conception:

Suppose $\left\{ξ_{n}\right\}$ are i.i.d. positive integer valued random variables. Let $R_{n}=|\{ξ_{1},…,ξ_{n}\}|$ denote the number of distinct elements among the first $n$ variables. Prove that $\lim  \mathbb{E} (R_{n}/n) = 0$.

I have tried to calculate $\mathbb{E} (R_{n}/n)$ directly with finite numbers of r.v.'s and a finite set of positive integers for r.v.'s to be valued in, and then take the limit. It didn't work. I also tried to employ some of the laws of large numbers, but couldn't find one suitable. I am not sure should I employ some analysis techniques such as truncation, etc. ? Or is there any other way out?
There is a related question, but I believe it is a weaker version of the above one:
What is the probability space/measure here? In general, what does "almost sure" convergence mean when a probability space is not explicitly specified?
 A: First observe that for a fixed $\omega$, $R_{n+1}(\omega)=R_n(\omega)$ if $X_{n+1}(\omega)$ belongs to $\{X_k(\omega),1\leqslant k\leqslant n\}$ and
$R_{n+1}(\omega)=R_n(\omega)+1$ otherwise. Therefore,
$$
R_{n+1}=R_n+\mathbf{1}\left(\bigcap_{i=1}^n\{X_{n+1}\neq X_i\}\right).
$$
Taking the expectations gives
$$
\mathbb E\left[R_{n+1}\right]=\mathbb E\left[R_{n}\right]+\mathbb P\left(\bigcap_{i=1}^n\{X_{n+1}\neq X_i\}\right).
$$
In order to compute the last probability, condition on the set $\{X_{n+1}=\ell\}$, $\ell\in\mathbb N$, more precisely,
$$
\mathbb P\left(\bigcap_{i=1}^n\{X_{n+1}\neq X_i\}\right)
=\sum_{\ell\in\mathbb N}\mathbb P\left(\bigcap_{i=1}^n\{\ell\neq X_i\}\cap\{X_n=\ell\}\right).
$$
Denoting $p_\ell:=\mathbb P(X_1=\ell)$, we get from independence and the fact that $(X_i)_{i\geqslant 1}$ is i.i.d. that
$$
\mathbb P\left(\bigcap_{i=1}^n\{X_{n+1}\neq X_i\}\right)
=\sum_{\ell\in\mathbb N} p_\ell\left(1-p_\ell\right)^n.
$$
Finally, we derive that
$$
\frac 1N\mathbb E\left[R_N\right]=\frac 1N\mathbb E\left[R_1\right]
+\sum_{\ell\in\mathbb N} p_\ell\frac{1}{N}\sum_{n=1}^N\left(1-p_\ell\right)^n
$$
and conclude by dominated convergence.

We got that
$$
\frac{R_N}N=\frac{R_1}N+\sum_{n=1}^N\mathbf{1}\left(\bigcap_{i=1}^n\{X_{n+1}\neq X_i\}\right)
$$
hence $R_N/N$ can be expressed as an average but the involved random variables are not independent. Maybe law of large numbers for martingales may be useful here, but at least a direct computation of the expectation does the job.
A: For each $i\in \mathbb N$, let $E^n_k$ be the event that $k$ appears in $\{\xi_1,\dots,\xi_n\}$. Obviously, $P(E^n_k)=1-(1-P(\xi_1=k))^n$. It follows by linearity of expectation that
$$
E[R_n/n]=\sum_{k=1}^\infty \frac{1-(1-P(\xi_1=k))^n}{n}
$$
Each summand looks like $\frac{1-(1-x)^n}{n}$. Using the inequality $(1-x)^n\ge 1-nx$, valid for $x\in [0,1]$, we have that
$$
 \frac{1-(1-P(\xi_1=k))^n}{n}\le P(\xi_1=k)
$$
Therefore, we can apply the dominated convergence theorem to show
$$
\lim_{n\to\infty} E[R_n/n]
=\lim_{n\to\infty}\sum_{k=1}^\infty  \frac{1-(1-P(\xi_1=k))^n}{n}
\,\,\stackrel{\text{DCT}}=\,
\sum_{k=1}^\infty  \lim_{n\to\infty}\frac{1-(1-P(\xi_1=k))^n}{n}=\sum_{k=1}^\infty 0=0
$$
